# There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and the third is also a biased coin that comes up tails 40% of the time. One of the three coins is chosen at random and tossed and it shows heads. What is the probability that it was the two-headed coin? - Mathematics

There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and the third is also a biased coin that comes up tails 40% of the time. One of the three coins is chosen at random and tossed and it shows heads. What is the probability that it was the two-headed coin?

#### Solution

Let E1 be the event of selecting the two-headed coin,
E2 be the event of selecting the biased coin that comes up heads 75% of the times,
E3 be the event of selecting the biased coin that comes up tails 40% of the times
and A be the event of getting head on the coin.

Then,

P(E_1)=P(E_2)=P(E_3)=1/3
P(A/E1) = Probability of getting a head on the coin, given that the coin is two-headed.

⇒ P(A/E_1)=1

P(A/E2) = Probability of getting a head on the coin, given that the coin is a biased coin that comes up heads 75% of the time.

⇒ P(A/E_2)=75/100=3/4

Also, P(A/E3) = Probability of getting a head on the coin, given that the coin is a biased coin that comes up tails 40% of the time.

⇒ P(A/E_3)=60/100=3/5

By Baye's theorem,

required probability = P(E1/A)

=( P(E_1) P(A/E_1))/(P(E_1) P(A/E_1)+P(E_2) P(A/E_2)+P(E_3) P(A/E_3))

= (1/3xx1)/(1/3xx1 )+ (1/3xx3/4) + (1/3xx3/5)

= 20/47

Thus, the probability that it was the two-headed coin is 20/47.

Is there an error in this question or solution?